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\(\left(12\cdot15-x\right)\cdot\dfrac{1}{4}=120\cdot\dfrac{1}{4}\)
\(\Rightarrow\left(180-x\right)\cdot\dfrac{1}{4}=30\)
\(\Rightarrow180-x=30:\dfrac{1}{4}\)
\(\Rightarrow180-x=120\)
\(\Rightarrow x=180-120\)
\(\Rightarrow x=60\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=1.2+2.3+3.4+...+n\left(n+1\right)\\ \Rightarrow3.C=1.2.3+2.3.3+3.4.3+..+n\left(n+1\right).3\\ \Rightarrow3.C=1.2.3+2.3.4-1.2.3+....+n\left(n+1\right)\left(n+2\right)-\left(n-1.n.\left(n+1\right)\right)\\ \Rightarrow3.C=n\left(n+1\right)\left(n+2\right)\\ \Rightarrow C=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
Cái D tính TT
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Ta có:\(\frac{1}{3x4}+\frac{2}{4x6}+\frac{3}{6x9}+\frac{4}{9x13}+\frac{5}{13x18}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)
\(=\frac{1}{3}-\frac{1}{18}=\frac{6}{18}-\frac{1}{18}=\frac{5}{18}\)
đặt \(A=\frac{1}{3.4}+\frac{2}{4.6}+\frac{3}{6.9}+\frac{4}{9.13}+\frac{5}{13.18}\)=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}=\frac{1}{3}-\frac{1}{18}=\frac{5}{18}\)
=>A=5/18
vậy........
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a) \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2.3\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\dfrac{\left(0,2\right)^5.3^5}{\left(0,2\right)^5.\left(0,2\right)}=\dfrac{3^5}{0,2}=\dfrac{243}{0,2}=1215\)
c) \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2=2:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^2=2:\left(-\dfrac{1}{6}\right)^2=2:\dfrac{1}{36}=72\)
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\(\frac{3\cdot4+3\cdot7}{6\cdot5+9}=\frac{3\left(4+7\right)}{2\cdot3\cdot5+3\cdot3}=\frac{3\cdot11}{3\left(2\cdot5+3\right)}=\frac{3\cdot11}{3\cdot13}=\frac{11}{13}\)
\(\frac{6\cdot9-2\cdot17}{63\cdot3-119}=\frac{2\cdot3\cdot9-2\cdot17}{7\cdot9\cdot3-7\cdot17}=\frac{2\left(27-17\right)}{7\left(27-17\right)}=\frac{2}{7}\)
\(\frac{21\cdot3-21}{1-24}=\frac{21\left(3-1\right)}{-23}=\frac{21\cdot2}{-23}=\frac{42}{-23}=-\frac{42}{23}\)
\(\frac{9\cdot3-54}{18}=\frac{9\cdot3-9\cdot6}{9\cdot2}=\frac{9\cdot\left(3-6\right)}{9\cdot2}=\frac{3-6}{2}=-\frac{3}{2}\)
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A=1.(2+2)+2.(2+3)+3.(2+4) + ... + 99.(2+100)
A=1.2+1.2+2.2+2.3+3.2+3.4+...+99.2+99.100
A=(1.2+2.2+3.2+4.2+...+99.2)+(1.2+2.3+3.4+...+99.100)
A=2.(1+2+3+4+...+99) + (1.2+2.3+3.4+...+99.100)
Tính B= 1.2+2.3+3.4+...+99.100 suy ra 3B= 1.2.3+2.3.3+3.4.3+...+99.100.3
3B= 1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3B=99.100.101
B= 33.100.101
A=2.99. (1+99):2+33.100.101
A=99.100+33.100.101=343200
![](https://rs.olm.vn/images/avt/0.png?1311)
2+3/2+4/3+5/4=73/12
KẾT QUẢ: \(\frac{73}{12}\)
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